A hydraulic automobile lift is designed to lift a vehicle of 4000kg. the area of cross section of the cylinder carrying the load is $250 c m^{2}$ . The maximum pressure the smaller piston would have to bear is (g=10m/s² )

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$1.6 \times 10^{6} p a$

$2 \times 10^{6} p a$

$200 \times 10^{6} p a$

$16 \times 10^{6} p a$

answer is A.

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Detailed Solution

$P = \frac{F_{2}}{A_{2}} = \frac{m g}{A_{2}} = \frac{4000 \times 10}{250 \times 10^{- 4}} = \frac{400}{25} \times 10^{5} = 16 \times 10^{5} = 1.6 \times 10^{6} p a$

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